# Compatibility of osculating elements and cartesian vectors given by JPL Horizons?

I'm facing a problem I can't explain and I would like to know if someone has more information about it.

I'm drawing a solar system and when I plot earth with the Horizon cartesian coordinates, it does not match the orbit plotted via the Horizon orbital elements.

Here is the query of the orbital elements

2458055.500000000 = A.D. 2017-Oct-29 00:00:00.0000 TDB
EC= 1.693936505240274E-02 QR= 9.828535727121969E-01 IN= 1.619499879517213E-03
OM= 1.180185004251146E+02 W = 3.460022671282024E+02 Tp=  2458123.158659660257
N = 9.859206292830410E-01 MA= 2.932939316914255E+02 TA= 2.914962922586066E+02
A = 9.997893698231427E-01 AD= 1.016725166934089E+00 PR= 3.651409548675242E+02


I'm using EC and A to plot the ellipse. Here are my calculations : Center of ellipse X : -EC * A Center of ellipse Y : 0 Radius of ellipse X : A Radius of ellipse Y : A * sqrt(1 - EC^2)

Here is the query of the cartesian coordinates

And here is the answer :

2458055.500000000 = A.D. 2017-Oct-29 00:00:00.0000 TDB
X = 8.085189297377026E-01 Y = 5.770742450594002E-01 Z =-2.783714659030374E-05
VX=-1.027876320855459E-02 VY= 1.393441165141858E-02 VZ= 7.146303246417249E-08
LT= 5.737033579585715E-03 RG= 9.933365717771512E-01 RR=-2.711916121500719E-04


I'm plotting earth at the given X, Y, Z coordinates

And I don't understand the result. Earth is almost on its orbit but not exactly. It's off by 1,275... % and I don't understand why.

Is it because the keplerian elements are approximations or is it because my drawing of the orbit is not very well made atm ? In other words, should the earth be exactly on its orbit when using the Horizon data or is this error expected ?

Thank you very much

• At a guess, you need to rotate the ellipse according to the argument of periapsis (W in the Horizon elements); the X axis is aligned to the ascending node of the orbit, not the axis of eccentricity. en.wikipedia.org/wiki/Argument_of_periapsis – Russell Borogove Oct 30 '17 at 0:56
• Thank you that was that. I added W and OM to get the right angle and now my orbit is passing almost in the middle of earth. At least it's going through it ! I'll try to improve it but this is already much better. – DBCL Oct 30 '17 at 21:49
• Check the chosen coordinate system in 'Table settings'. There are 3 options, you probably need 'Earth mean equator and equinox of reference epoch' – Tarlan Mammadzada Mar 13 '18 at 11:06
• There may be a slip in the OP's definition of the ellipse in the X-Y plane. If the radius has x-coord A, then the center is at x-coord 0, the quoted x-coord -EC*A belongs to a focus, not the center. The calculations weren't given, but for an ellipse with eccentricity around 0.0169, inadvertent substitution of a focus for the center could account for a discrepancy in radius coordinate of the order 1--2%. – terry-s Mar 13 '18 at 23:22

You must use all six the orbital parameters. @RussellBorogove points out one, but $e, a, i, \Omega, \omega, L$ are all needed to obtain the position of an object at a single point in time in three dimensions.

There are several other answers here in SXSE that explain how to get the position in 3D from these parameters. Check the following; 1, 2, 3. 4, 5 and 6, listed in no particular order.

As an asside, Keplerian osculating orbits are handy, but the reality of gravity is that in a complex system like this — Earth having such a big moon, the Sun not fixed because of Jupiter and the other giant planets — they are just approximations. It's why the numbers change daily, and it's why if you use these osculating elements to predict a position at any other time, even for example half way between one day and the next in your example, you'll get a small error.

below: Vector diagram of Encke's Method of perturbation analysis, showing the osculating orbit, the perturbed orbit, and a perturbing body. From here.

• Projecting the orbital ellipse into 2D, and with Earth's current position as a given, you don't need all the elements to get things to line up the way OP is trying to do. – Russell Borogove Oct 30 '17 at 2:19
• @RussellBorogove then I'll encourage you to post an answer and show how it's done and what residual error is still left. That's not an answer I'd be comfortable posting, but I'll up vote yours if you do. I suppose if the idea is to only get within 1% of 1 AU it might be sufficient. Remember though that you can't propagate this orbit because it pretends that the Moon does not exist; the OP is using the instantaneous motion of the Earth's geocenter, not that of the Earth-Moon barycenter. – uhoh Oct 30 '17 at 4:04
• OP isn’t propagating anything. – Russell Borogove Oct 30 '17 at 4:11
• @RussellBorogove yep, it's just good to include that in any answer lest anyone think about trying. – uhoh Oct 30 '17 at 4:19
• Thank you for your detailled answer and for the informations about OP, but I don't want to plot earth position in cartesian coordinates based on keplerian elements. @Russell 's suggestions put me in the right direction. – DBCL Oct 30 '17 at 21:51